Question

Dy/dx = (sin x)/y , y(0) = 2

Answer 1

Answer:

The solution for this differential equation is $y=\sqrt{-2cos(x)+6}$

Step-by-step explanation:

This differential equation $\frac{dy}{dx}=\frac{sin(x)}{y}$ is a separable First-Order ordinary differential equation.

We know this because a first-order differential equation is separable if and only if it can be written as

$\frac{dy}{dx}=f(x)g(y)$ where f and g are known functions.

And we have

$\frac{dy}{dx}=\frac{sin(x)}{y}\\ \frac{dy}{dx}=sin(x)\frac{1}{y}$

To solve this differential equation we need to integrate both sides

$y\cdot dy=sin(x)\cdot dx\\ \int\limits {y\cdot dy}= \int\limits {sin(x)\cdot dx}$

$\int\limits {y\cdot dy}=\frac{y^{2} }{2} + C$

$\int\limits {sin(x) \cdot dx}=-cos(x) + C$

$\frac{y^{2} }{2} + C=-cos(x) + C$

We can make a new constant of integration $C_{1}$

$\frac{y^{2} }{2}=-cos(x) + C_{1}$

We need to isolate y

$\frac{y^{2} }{2}=-cos(x) + C_{1}\\y^2=-2cos(x)+2*C_{1}\\\mathrm{For\:}y^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}y=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\y=\sqrt{-2cos(x)+c_{1} } \\y=-\sqrt{-2cos(x)+c_{1} }$

We have the initial conditions y(0)=2 so we can find the value of the constant of integration for $y=\sqrt{-2cos(x)+c_{1} }$

$2=\sqrt{-2\cos \left(0\right)+c_1}\\2= \sqrt{-2+c_1} \\c_1=6$

For $y=-\sqrt{-2cos(x)+c_{1} }$ there is not solution for $c_{1}$ in the domain of real numbers.

The solution for this differential equation is $y=\sqrt{-2cos(x)+6}$